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ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2009/a/1/b/solution

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Past exam of the approximation theory and finite element methods course of the University of Oxford / 2009 / A / 1 / b / Solution by

Codex  0 2026-10-31

With

w_{s} = 1/ π_{n, s} (x_{s})

, the Lagrange interpolation polynomial can be written, away from its nodes,

p_{n} (x) = π_{n} (x) \sum_{s = 0}^{n} \frac{w _{s} u _{s}}{x - x _{s}} .

Apply the same formula to the constant data

u_{s} = 1

. Uniqueness of polynomial interpolation gives

\sum_{s} ℓ_{s} = 1

, and hence

1 = π_{n} (x) \sum_{s = 0}^{n} \frac{w _{s}}{x - x _{s}} .

Divide the two identities to obtain the barycentric interpolation formula

p_{n} (x) = \frac{s = 0 \sum n w _{s} u _{s} / ( x - x _{s} )}{s = 0 \sum n w _{s} / ( x - x _{s} )} .

At

x = x_{r}

, use the continuous limiting value

u_{r}

. Multiplying every weight by the same nonzero constant cancels between numerator and denominator, so only relative weights are needed in computation.

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